# Elements conjugate in profinite completion

Problem. Let $$G$$ be a residually finite group, and identify $$G$$ with its image under the canonical map to its profinite completion $$\hat{G}$$. Let $$x,y \in G$$. Prove that the following conditions are equivalents:

(i) $$x,y$$ are conjugate in $$\hat{G}$$.

(ii) the imagens of $$x,y$$ in $$G/K$$ are conjugate in $$G/K$$, for every normal subgroup $$K$$ of finite index in $$G$$.

We know that $$\hat{G} = \varprojlim_{K \in I}G/K$$ where $$I$$ is a non-empty filter base of normal subgroups of finite index in $$G$$ and there is a continuous homomorphism $$\varphi: G \to \hat{G}$$ given by $$\varphi(g) = Kg$$. The pair $$(\hat{G},\varphi)$$ has the property:

"if $$\psi: G \to H$$ is a continuous homomorphism, to a finite group $$H$$, then there is a unique homomorphism $$\overline{\psi}: \hat{G} \to H$$ such that $$\psi = \overline{\psi}\circ\varphi$$."

1. If $$x,y \in G$$, then $$Kx,Ky \in \hat{G}$$. So, the item (i) must be "... in $$G$$"?

2. Who are the imagens of $$x,y$$ in $$G/K$$? Since $$\hat{G} = \varprojlim G/K$$, then $$(G/K_{i},f_{ij})$$ is a inverse system with inverse limit $$(\hat{G},f_{i})$$ where $$f_{i}: \hat{G} \to G/K_{i}$$? We can connect $$G$$ to $$\hat{G}$$ by $$\varphi$$ and $$\hat{G}$$ to $$G/K_{i}$$ by $$f_{i}$$, but who is $$f_{i}$$? I don't know a definition of completion using, explicitly, inverse systems indexed by a direct set and its maps. Thus, I don't know how to start this problem.

• $\hat{G}$ is the set of sequences $(a_K K)_{K \in I}$ (indexed by finite index normal subgroups $K$) such that $a_K \in G, a_K K\in G/K$ and $a_K H = a_H H$ whenever $K\subset H$. Trying with $G = \Bbb{Z}$ and $I$ the index $p^n$ subgroups yields the $p$-adic integers, with $I$ every subgroup it is the profinite integers. – reuns Apr 21 at 20:28
• Just a word on your profile: if you're tackling problems like this, then, surely, you'll be able to help someone. Have confidence. You're doing well :) – Shaun Apr 21 at 20:51
• @Shaun thank you for the encouragement! – Lucas Corrêa Apr 21 at 21:04
• You can also think to $\hat{G}$ as the set of (limits of) sequences $(g_n)_{n \ge 1}$ such that $g_n K$ ends being constant for every $K$, then $a,b\in G$ are conjugate in $\hat{G}$ if $a^{-1} (g_n b g_n^{-1}) K \to K$ for every $K$. – reuns Apr 21 at 21:31
• $p$-adic integers $\Bbb{Z}_p$ are the completion of $\Bbb{Z}$ for the $p$-adic metric $d(b,b+a) = |a|_p = p^{-k}$ if $a \equiv 0 \bmod p^k,a \not \equiv 0 \bmod p^{k+1}$. Do you see how an equivalent metric can be defined in any group $G$ to obtain $\hat{G}$ as the completion ? – reuns Apr 21 at 21:50

You don't need a "filter basis of normal subgroups" to define $$\hat{G}$$. Take $$\hat{G}= \underset{\leftarrow}{\mathrm{lim}} \: G/K$$ over every normal subgroup of finite index $$K$$ of $$G$$. Then $$\hat{G}$$ is a subgroup of the product $$\prod_K G/K$$ (over the same $$K$$'s) as in the comment by @reuns, and one has projections $$p_K :\hat{G} \to G/K$$ for every normal subgroup of finite index $$K$$ of $$G$$, that are factorizations of the common projections $$\pi_K : G \to G/K$$ : $$\pi_K = G \overset{i}{\to} \hat{G} \overset{p_K}{\to} G/K$$ (where $$i$$ is the natural inclusion).
So $$(i) \Rightarrow (ii)$$ is pretty clear.
For the other inclusion, you have to build an element $$z=(z_K)_K \in \hat{G}$$ such that $$zxz^{-1}=y$$ from $$z_K$$'s such that $$z_K p_K(x) z_K^{-1}=p_K(y)$$ for every normal subgroup of finite index $$K$$ of $$G$$.